## Precalculus (6th Edition)

Published by Pearson

# Chapter 1 - Equations and Inequalities - Chapter 1 Test Prep - Review Exercises - Page 178: 43

#### Answer

The solution is $x=2\pm\sqrt{6}$

#### Work Step by Step

$(2x+1)(x-4)=x$ Evaluate the product on the left side: $2x^{2}-8x+x-4=x$ Take $x$ to the left side and simplify: $2x^{2}-8x+x-4-x=0$ $2x^{2}-8x-4=0$ Divide the whole equation by $2$: $\dfrac{1}{2}(2x^{2}-8x-4=0)$ $x^{2}-4x-2=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. In this case, $a=1$, $b=-4$ and $c=-2$. Substitute the known values into the formula and evaluate: $x=\dfrac{-(-4)\pm\sqrt{(-4)^{2}-4(1)(-2)}}{2(1)}=\dfrac{4\pm\sqrt{16+8}}{2}=...$ $...=\dfrac{4\pm\sqrt{24}}{2}=\dfrac{4\pm2\sqrt{6}}{2}=\dfrac{4}{2}\pm\dfrac{2\sqrt{6}}{2}=2\pm\sqrt{6}$ The solution is $x=2\pm\sqrt{6}$

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